11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer If the roots of the equations \[{{x}^{2}}-bx+c=0\] and \[{{x}^{2}}-cx+b=0\] differ by the same quantity, then  \[b+c\] is equal to                            [BIT Ranchi 1969; MP PET 1993]

    A) 4

    B) 1

    C) 0

    D) -4

    Correct Answer: D

    Solution :

    Let the roots are \[\alpha ,\beta \] of  \[{{x}^{2}}-bx+c=0\]and \[{\alpha }',{\beta }'\]be roots of \[{{x}^{2}}-cx+b=0\] Now \[\alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }=\sqrt{{{b}^{2}}-4c}\] .....(i) and \[\alpha '-\beta '=\sqrt{{{(\alpha '+\beta ')}^{2}}-4\alpha '\beta '}=\sqrt{{{c}^{2}}-4b}\] .....(ii) But \[\alpha -\beta =\alpha '-\beta '\] Þ \[\sqrt{{{b}^{2}}-4c}=\sqrt{{{c}^{2}}-4b}\,\,\Rightarrow \,\,{{b}^{2}}-4c={{c}^{2}}-4b\] Þ \[{{b}^{2}}-{{c}^{2}}=4c-4b\] Þ \[(b+c)(b-c)=4(c-b)\]Þ\[b+c=-4\]

You need to login to perform this action.
You will be redirected in 3 sec spinner